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Let $X$ be the sequence space of all sequences of the form $$ x =(x_1,x_2 ,x_3,\dots , x_n ,0,0 \dots ) $$ where $\forall i \in \mathbb{N}$ $x_i \in \mathbb{R}$

whose terms are all zero after some index

$(X, || .||_{\infty})$ is a Normed Space where $$||x||_{\infty} =\max _{i \in \mathbb{N}}|x_i|$$

Prove (or disprove ) if it is a complete norm space


Ammo

Def complete space seq in x that is cauchy then converges in X

Def Cauchy $$\forall \epsilon >0 , \exists N \in \mathbb{N} ; n,m \geq N \Rightarrow || x_n -x_m|| < \epsilon $$

Def of convegence in Normed space $\lim_{n \to \infty }|| x_n||=x$ $$\forall \epsilon >0 , \exists N \in \mathbb{N} ; n\geq N \Rightarrow || x_n -x|| < \epsilon $$

Prop from Lecture notes

Let $p \in \mathbb{N}, \mathbb{R}^p$ is complete normed vector space if it is equipped with $|| .||_{\infty}$


Tried to play with $$ \begin{aligned}\begin{pmatrix} \frac{1}{n} \\ 1 \\ 0 \\ \vdots\\0 \end{pmatrix}=x_n &&\text{and} &&x=\begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots\\0 \end{pmatrix} \end{aligned}$$

I know that $x_n$ is cauchy because ${1/n}$ is cauchy. and $$ x_n -x_m = \begin{pmatrix} \frac{1}{n} -\frac{1}{m}\\ 0\\ 0 \\ \vdots\\0 \end{pmatrix}$$ and that should be cauchy. $x_n-x$ seems to converge using the definition of convergence. because

$$x_n-x =\begin{pmatrix} 1/n \\ 0 \\\ 0\\\ \vdots \end{pmatrix}$$

Having trouble coming up with a sequence that is cauchy in X and does not convege in X.

Want to say that since $(R^p,|| .||_\infty)$ is complete and $X \subset R^p$ then $X$ is complete but dont know that can be said.


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  • $\begingroup$ There is also another (slightly more advanced) ammo which can be used to prove something even stronger: Theorem: The dimension of every Banach space is uncountable. Your space $X=c_{00}$ has a countable Hamel basis, so there is no norm on $X$ which can make it a complete space. $\endgroup$ Commented Sep 29, 2016 at 17:07

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For example $x_n= (1,1/2,\dots,1/n,0,\dots)$. Then $x_n$ is Cauchy, and $x_n\to x$ in $\ell_{\infty}$ sense where $x=(1,1/2,1/3,...,1/n,1/(n+1),\dots)$ but $x\notin X$.

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